Solve for x: x^4 + x^3 + x^2 + x + 1 = 0.
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If you draw the graph of x^4 + x^3 + x^2 + x + 1, it won't cut the x-axis. So, x^4 + x^3 + x^2 + x + 1 = 0 has no real roots
Thanks songoku, I have an UGLY solution for this, but i am hoping THAT someone may be able to solve this elegantly.
If anyone's curious, the four complex roots are the following: x ≈ 0.309 + 0.951i x ≈ 0.309 - 0.951i x ≈ -0.809 + 0.588i x ≈ -0.809 - 0.588i (using a computer, of course ) Polar form: 01
Originally Posted by pacman Solve for x: x^4 + x^3 + x^2 + x + 1 = 0. Observe that Hence the solution you want is the set of the four complex 5th roots of unity. No need to plot graph or use computer.
Hint: divide BS x^2, we have x^2 + x + 1 + 1/x + 1/x^2 = 0, rearranging it and then grouping it (x + 1/x)^2 + (x + 1/x) + 1 = 0 It is now quadratic in x + 1/x, the result is quite ugly but it can generate a result . . .
Originally Posted by pacman Hint: divide BS x^2, we have x^2 + x + 1 + 1/x + 1/x^2 = 0, rearranging it and then grouping it (x + 1/x)^2 + (x + 1/x) + 1 = 0 It is now quadratic in x + 1/x, the result is quite ugly but it can generate a result . . . Then Is this what you are saying?
I think there's a typo in pacman's last post. Instead of it should have been 01
i mixed it up in a bowl of soup, yeongil is correct with his observation . . . .
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